## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x \to -1}(x^4-3x)(x^2+5x+3)=-4$
$\lim\limits_{x \to -1}(x^4-3x)(x^2+5x+3)$ $=\lim\limits_{x \to -1}(x^4-3x)\times\lim\limits_{x \to -1}(x^2+5x+3)$ (product law) $=[\lim\limits_{x \to -1}x^4-\lim\limits_{x \to -1}3x]\times[\lim\limits_{x \to -1}x^2+\lim\limits_{x \to -1}5x+\lim\limits_{x \to -1}3]$ (difference and addition law) $=[\lim\limits_{x \to -1}x^4-3\lim\limits_{x \to -1}x]\times[\lim\limits_{x \to -1}x^2+5\lim\limits_{x \to -1}x+\lim\limits_{x \to -1}3]$ (constant multiple law) $=[(-1)^4-3\times(-1)]\times[(-1)^2+5\times(-1)+3]$ $=-4$